# Program to find longest even value path of a binary tree in Python

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Suppose we have a binary tree, we have to find the longest path consisting of even values between any two nodes in the tree.

So, if the input is like

then the output will be 5 as the longest path is [10, 2, 4, 8, 6].

To solve this, we will follow these steps −

• ans := 0

• Define a function find() . This will take node

• if node is null, then

• return (-1, -1)

• leftCnt := maximum of returned value of find(left of node) + 1

• rightCnt := maximum of returned value of find(right of node) + 1

• if value of node is even, then

• ans := maximum of ans and (leftCnt + rightCnt + 1)

• return (leftCnt, rightCnt)

• otherwise,

• ans := maximum of ans, leftCnt and rightCnt

• return (-1, -1)

• From the main method do the following −

• find(root)

• return ans

Let us see the following implementation to get better understanding −

## Example

class TreeNode:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def solve(self, root):
ans = 0
def find(node):
nonlocal ans
if not node:
return -1, -1
leftCnt = max(find(node.left)) + 1
rightCnt = max(find(node.right)) + 1
if node.val % 2 == 0:
ans = max(ans, leftCnt + rightCnt + 1)
return leftCnt, rightCnt
else:
ans = max(ans, max(leftCnt, rightCnt))
return -1, -1
find(root)
return ans
ob = Solution()
root = TreeNode(2)
root.left = TreeNode(10)
root.right = TreeNode(4)
root.right.left = TreeNode(8)
root.right.right = TreeNode(2)
root.right.left.left = TreeNode(6)
print(ob.solve(root))

## Input

root = TreeNode(2)
root.left = TreeNode(10)
root.right = TreeNode(4)
root.right.left = TreeNode(8)
root.right.right = TreeNode(2)
root.right.left.left = TreeNode(6)

## Output

5
Updated on 10-Oct-2020 12:42:34